of the case, and are of such size that the outer edges just clear the inner circumference of the case on top and the rib E on the bottom, while the ends just clear the inside of the end plates of the case. The shaft around which the blades rotate is bent down at each end, beyond the blades, and passes through the hollow shaft of the drum to an outboard bearing on the pulley end, while at the other end it projects just beyond the outside of the bearing in the case. As the drum rotates, its circumference, at the points where the blades pass through it, is continually changing its position relatively to the blades, and the function of the rolls, R, R, R, is to permit of a continual sliding motion at these points by rocking back and forth in their bearings in the drum, so as to keep the slots through which the blades pass always in line with the blades. The action of the exhauster is as follows: The drum, D, being rotated in the direction of the arrow, carries the blades, B, B, B, with it, and these blades being practically in contact with the circumference and ends of the case, force the gas ahead of them, and thus pass it from inlet to outlet. As the drum is practically in contact with the rib, E, the gas driven forward on top cannot pass back on the bottom, and must go out by the outlet. By a study of the cross-section it will be seen that in this, as in all similar forms of rotary exhausters, the space on one side is continually increasing in volume, while that on the other is continually decreasing, thus drawing in gas on the one side, the inlet, and forcing it out on the other side, the outlet. Outlet Inlet Fig. 2. Cross Section of Root Exhauster. A vertical cross-section of the Root exhauster is shown on the accompanying cut. F, is a cast-iron case consisting of two half cylinders of equal diameter, separated by a rectangular prism, and with flat ends. The moving parts consist of two impellers, C and D, keyed respectively to the shafts, A and B, and so set in the case that the axes of the corresponding shaft, impeller and half cylinder coincide. The impellers are so shaped that as they rotate each is always in contact, at some point, with the cylindrical portions of the case, and at the same time in contact with the other, while the ends are practically in contact with the ends of the case. The shafts are supported by bearings on the case, and connected with each other at each end by two gear wheels, one on each shaft, of the same diameter, and with an equal number of teeth, so that the relative positions of the impellers is always the same, at the same point in any revolution. The lower shaft is extended at one end and provided with a belt pulley, or connected direct to an engine on the same bed plate as the exhauster. The action is as follows: Motion being imparted to shaft B, is transmitted through the gear wheels to shaft A, causing the two impellers to rotate, as shown by the arrows and by the continual increase in volume of the space on the inlet side and decrease in volume of that on the outlet side, to draw the gas from the inlet and force it through the outlet, the contact. between the impellers preventing the gas from passing back through the middle. (Trustees.) 5. Using the weights per cubic foot of the various gases. given in the answer to Question No. 6, Second Series, and the definition of specific gravity given in the answer to Question No. 5, Second Series, calculate (1) the weight of 1,000 cu. ft. of coal gas with a specific gravity of .430; (2) the weight of 1,000 cu. ft. of water gas with a specific gravity of .640; (3) the specific gravity of marsh gas (CH1), and (4) the specific gravity of ethylene (C2H ̧). Give your calcu lations in each case. Ans. The specific gravity of a gas is defined in the answer to Question No. 5 of the Second Series as the ratio of the weight of any given volume of the gas to the weight of an equal volume of air, the air and gas being measured at the same temperature and pressure. The weight of 1,000 cu. ft. of coal gas with a specific gravity of .430 will therefore be equal to the weight of 1,000 cu. ft. of air multiplied by .430. As the weight of 1 cu. ft. of air at 60° F. and 30 in. pressure is given in the answer to Question No. 6 of the Second Series as .076357 lbs., 1,000 cu. ft. will weigh 76.357 lbs. Hence, 76.357 X.430=32.83 lbs. is the weight of 1,000 cu. ft. of coal gas with a specific gravity of .430. In the same way the weight of 1,000 cu. ft. of water gas with a specific gravity of .640 is equal to 76.357X.640=48.87 lbs. The specific gravity of marsh gas according to the definition will be equal to the quotient obtained by dividing the weight of a cubic foot of marsh gas by that of a cubic foot of air. The weight of marsh gas is given as .042308, therefore the specific gravity of marsh gas equals .55408. In the same way the specific gravity of ethylene is equal to .076357 .96964. (Trustees.) .042308 = .074039 6. In a gas pipe, one end of which is higher than the other, such as the riser pipe in a house or a street main running up hill, and in which there is no flow of gas for the time being, the pressure as shown by a syphon pressure gauge will not be the same at the top and at the bottom. At which point will it be the greater? Does the specific gravity of the gas have any influence upon the amount of this difference between the pressure at the two points? Give the reasons for your answers. Ans. A syphon gauge will show a greater pressure at the high end of a pipe filled with illuminating gas through which there is at the time no flow of gas, than it will show at the low end. The amount of the difference between the pressure at the high end and that at the low end is influenced by the specific gravity of the gas and increases as the specific gravity decreases. The existence of this difference in gauge pressure between the top and the bottom of an inclined pipe and the dependence of its amount upon the specific gravity of the gas are due to the facts that the syphon gauge does not show actual pressures, but only the difference between the actual pressures exerted on each leg of the gauge, and that the actual pressure of both the outside air and of gas at rest in a pipe decreases in proportion to their weight as the distance of the point of observation from the centre of the earth increases. The actual pressure of both air and gas will be less at the top of the pipe than at the bottom by an amount equal to the weight of a column of air or gas of unit area and of a height equal to the difference in elevation between the two ends of the pipe. As the specific gravity of illuminating gas is less than one, that is, such gas is lighter than air, the weight of the column of gas will be less than that of the column of air. Therefore, the actual pressure of the gas will decrease less than the actual pressure of the air, and consequently the difference between the actual pressure of the gas and that of the air will increase with an increase in elevation. The lighter the gas, that is, the lower its specific gravity, the smaller will be the decrease in its absolute pressure and the greater the increase in the gauge pressure for any given increase in elevation. The amount of this increase can be obtained for any particular case by dividing the difference between the weight. of a column of air with an area of one square inch and a height equal to the difference in elevation between the two ends of the pipe and that of a column of gas of the same area and height by the weight of a cubic inch of water. For a gas with a specific gravity of .700 and for a difference of elevation of 100 ft. between the two ends of the pipe, the calculation would be as follows: The height of the columns of air and gas will be 100 ft. or 1,200 in., and as the area of the columns is one square inch, the volume will be 1,200 cubic inches. A cubic foot of air 1728 10 weighs .076357 lb., and a cubic inch weighs 076357 =.00004419. 1,200 cubic inches of air, then, weigh .00004419 X 1,200 = .053028 lb. Since the specific gravity of the gas is .700, the column of gas will weigh as much as the column of air, and the difference between the weight of the column of air and that of the column of gas will be equal to the weight of the column of air, or to .053028X.3=.0159084. The weight of a cubic inch of water is .036 lb. and the increase in pressure will therefore be .44 in. Thus, a gas with a specific gravity of .700 will show an increase in pressure of .044 in. for each 10 feet increase in elevation, or an increase in pressure of in. for each 22.7 ft. of rise. .0159084 In the same way we find that a gas with a specific gravity of .600 will show an increase in pressure of .059 in. for each 10 ft. of rise, or an increase of in. pressure for each 16.9 ft. of rise; gas with a specific gravity of .500 will show an increase of .074 in. for each 10 ft. of rise, or an increase of in. for each 13.5 ft. of rise; gas with a specific gravity of .400 will show an increase of .088 in. for each 10 ft. of rise, or in. for each 11.4 ft. of rise, and gas with a specific gravity of .300 |